Problem: Factor completely. $64x^2-144x+81=$
Both $64x^2$ and $81$ are perfect squares, since $64x^2=({8x})^2$ and $81=({9})^2$. Additionally, $144x$ is twice the product of the roots of $64x^2$ and $81$, since $144x=2({8x})({9})$. $64x^2-144x+81 = ({8x})^2-2({8x})({9})+({9})^2$ So we can use the square of a difference pattern to factor: ${a}^2 -2( a)( b)+ {b}^2 =({a}-{b})^2$ In this case, ${a}={8x}$ and ${b}={9}$ : $ ({8x})^2-2({8x})({9})+({9})^2 =({8x}-{9})^2$ In conclusion, $64x^2-144x+81=(8x -9)^2$ Remember that you can always check your factorization by expanding it.